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Chapter 18: Problem 75

(a) Explain why Mg \(\mathrm{Mg}(\mathrm{OH})_{2}\) precipitates when \(\mathrm{CO}_{3}^{2-}\) ion is added to a solution containing \(\mathrm{Mg}^{2+} .\) (b) Will Mg \((\mathrm{OH})_{2}\) precipitate when 4.0 \(\mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is added to 1.00 \(\mathrm{L}\) of a solution containing 125 \(\mathrm{ppm}\) of \(\mathrm{Mg}^{2+} ?\)

Short Answer

Expert verified
(a) When CO鈧兟测伝 ion is added to a solution containing Mg虏鈦, it forms soluble MgCO鈧, decreasing the concentration of Mg虏鈦 ions in the solution and shifting the equilibrium of Mg(OH)鈧 dissolution reaction, causing more Mg(OH)鈧 to precipitate. (b) After calculating the ion product ([Mg虏鈦篯[OH鈦籡虏) and comparing it to the solubility product (Ksp) of Mg(OH)鈧, we find that the ion product (5.45 脳 10鈦宦孤) is very close to but still less than the Ksp (5.61 脳 10鈦宦孤), so Mg(OH)鈧 precipitation will not occur.

Step by step solution

01

Part (a): Explain why Mg(OH)鈧 precipitates when CO鈧兟测伝 is added to a solution containing Mg虏鈦

Adding CO鈧兟测伝 ion to a solution containing Mg虏鈦 ion results in the formation of MgCO鈧, which is soluble in water. Once Mg虏鈦 ions react with CO鈧兟测伝 ions, an equilibrium is established between Mg(OH)鈧 and MgCO鈧, as well as water and OH鈦 ions. The chemical reactions can be represented as follows: Mg(OH)鈧(s) 鈬 Mg虏鈦(aq) + 2OH鈦(aq) Mg虏鈦(aq) + CO鈧兟测伝(aq) 鈬 MgCO鈧(aq) When CO鈧兟测伝 ions are added to the solution, some Mg虏鈦 ions will react with CO鈧兟测伝 ions to form soluble MgCO鈧. This decreases the concentration of Mg虏鈦 ions in the solution, and as a result, the equilibrium shifts leftward in the first reaction (Le Chatelier's principle), causing more Mg(OH)鈧 to precipitate.
02

Part (b): Will Mg(OH)鈧 precipitate when 4.0 g of Na鈧侰O鈧 is added to 1.00 L of a solution containing 125 ppm of Mg虏鈦?

To determine if Mg(OH)鈧 will precipitate, we need to compare the ion product ([Mg虏鈦篯[OH鈦籡虏) with the solubility product (Ksp) for Mg(OH)鈧. If the ion product is greater than Ksp, then precipitation will occur; otherwise, it will not. Step 1: Calculate the concentrations of Mg虏鈦 and CO鈧兟测伝 in the solution Mg虏鈦 concentration = 125 ppm = 125 mg/L To convert this to mol/L, divide by the molar mass of Mg虏鈦, which is 24.3 g/mol: Mg虏鈦 concentration = \( \dfrac{125 \times 10^{-3}}{24.3} \) mol/L 鈮 0.00514 mol/L Na鈧侰O鈧 mass = 4.0 g To find the CO鈧兟测伝 concentration, divide by the molar mass of Na鈧侰O鈧 (105.99 g/mol) and by the volume of the solution in liters (1 L): CO鈧兟测伝 concentration = \( \dfrac{4.0}{105.99} \) mol/L 鈮 0.0377 mol/L Step 2: Calculate the concentration of OH鈦 produced by the reaction between CO鈧兟测伝 and Mg虏鈦 Mg虏鈦 + CO鈧兟测伝 鈬 MgCO鈧 + 2OH鈦 Since 1 mol of Mg虏鈦 reacts with 1 mol of CO鈧兟测伝, the limiting reactant is Mg虏鈦 (0.00514 mol/L), and the resulting concentration of OH鈦 is twice the limiting reactant (2 x 0.00514 mol/L) 鈮 0.0103 mol/L. Step 3: Calculate the ion product and compare it to the Ksp of Mg(OH)鈧 Ksp (Mg(OH)鈧) = 5.61 脳 10鈦宦孤 Ion product = [Mg虏鈦篯[OH鈦籡虏 = (0.00514)(0.0103)虏 鈮 5.45 脳 10鈦宦孤 Since the ion product is very close to but still less than the Ksp, Mg(OH)鈧 precipitation will not occur.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept that helps us understand how a chemical system at equilibrium responds to changes in concentration, pressure, volume, or temperature. When a change is imposed on a dynamic equilibrium system, the system will shift its position to counteract the effect of the change, thus re-establishing equilibrium.
This principle is particularly evident in chemical reactions involving equilibrium states. When addition of external ions or reactants occurs, the system adjusts by shifting the equilibrium position to reduce the effect of the added components.
For example, in the case of adding \(\text{CO}_3^{2-}\) ions to a solution containing \(\text{Mg}^{2+}\), the formation of \(\text{MgCO}_3\) occurs, which is more soluble. As \(\text{Mg}^{2+}\) from the solution gets consumed, the equilibrium in the reaction \(\text{Mg(OH)}_2 \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-\) shifts to the left, producing more \(\text{Mg(OH)}_2\), hence causing the precipitation of magnesium hydroxide. This adjustment is a direct application of Le Chatelier's Principle in balancing the chemical reaction.
Solubility Product Constant
The Solubility Product Constant, or \(K_{sp}\), of a compound is a key concept in predicting whether a precipitate will form when two solutions are mixed. It is specific to sparingly soluble ionic compounds and represents the product of the concentrations of the ions in a saturated solution, raised to the power of their respective coefficients in the balanced chemical equation.
For example, for magnesium hydroxide, \(\text{Mg(OH)}_2\), the \(K_{sp}\) expression is given by \([\text{Mg}^{2+}][\text{OH}^-]^2\). The value of \(K_{sp}\) can predict if a compound will precipitate under certain conditions. If the ion product of the solutions exceeds the \(K_{sp}\), the solution is supersaturated, and precipitation will occur.
In the context of the exercise provided, we compare the calculated ion product of the solution to the given \(K_{sp}\) for magnesium hydroxide, which allows us to determine whether a precipitate will form in the solution when additional ions are introduced.
Precipitation Reaction
Precipitation reactions occur when ions in solution combine to form an insoluble compound, which separates as a solid. This process is pivotal in chemistry, especially in qualitative analysis to identify ions in a solution and in various industrial applications.
For a precipitation reaction to occur, the product of the reacting ion concentrations (ion product) in the solution must exceed the compound's solubility product constant, \(K_{sp}\).
Consider the reaction in the given problem where \(\text{Mg(OH)}_2\) precipitates when \(\text{CO}_3^{2-}\) ions are added to a solution of \(\text{Mg}^{2+}\). Here, the interaction between \(\text{Mg}^{2+}\) and \(\text{OH}^-\) ions leads to the formation of solid \(\text{Mg(OH)}_2\). This approach helps predict the conditions under which precipitation happens, serving as a practical application of ionic equilibrium in chemical reactions.
Ion Product
The Ion Product is a crucial concept in determining whether a solution will undergo precipitation. It is calculated from the concentrations of the ions in solution, similar to the solubility product constant (\(K_{sp}\)). However, while \(K_{sp}\) is a constant specific to a particular substance at a given temperature, the ion product varies with the concentrations of dissolved ions.
When the ion product exceeds \(K_{sp}\), the solution is supersaturated, prompting a shift towards the formation of a solid precipitate as the reaction seeks equilibrium. If the ion product is less than \(K_{sp}\), the solution remains unsaturated, and no precipitation occurs.
In the example from the exercise, determining the ion product by calculating \([\text{Mg}^{2+}][\text{OH}^-]^2\) shows whether \(\text{Mg(OH)}_2\) will precipitate when \(\text{Na}_2\text{CO}_3\) is added. Since the calculated ion product was slightly less than the \(K_{sp}\), no precipitation occurs, highlighting the importance of understanding ion product in predicting chemical behavior in solutions.

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